[SPLIT, MOVED] number puzzles

[wrb]

Will you help me solve this code A B B
-A
D D
____________

 

[wrb]

Re: IMP 3...POW 7... Crack the Code...HELP QUICK PLEASE!!!

Hello

I have 3 codes I need help with can u also explain them throughly ...

1
A B B
-A
_____
D D



2
E E
+F F
_____
E E K



3
P A
+A P
_____
P U G

THANKS

 

[wrb]

Re: IMP 3...POW 7... Crack the Code...HELP QUICK PLEASE!!!

How do I get my answer and how loong does it take

 

[Gene]

ABB-A=DD or
DD+A=ABB
Since you are adding a single digit to a two digit number there can only be one to carry so A = 1.
Now you have
DD+1=1BB
The only digit you can add 1 to and get a carry is 9 so D=9
That's the way you do it. If you need more help, show me how far you have gotten on the others.

 

[wrb]

ABB-A=DD or
DD+A=ABB
Since you are adding a single digit to a two digit number there can only be one to carry so A = 1.
Now you have
DD+1=1BB
The only digit you can add 1 to and get a carry is 9 so D=9
That's the way you do it. If you need more help, show me how far you have gotten on the others.

 

[soroban]

Re: IMP 3...POW 7... Crack the Code...HELP QUICK PLEASE!!!

Hello, wrb!

These are often called "Alphametics"
. . requiring familiarity with simple arithmetic.

\(\displaystyle 1)\;\;\begin{array}{ccc} A\;B\;B \\ - \quad A \\ \hline \;\;\;D\; D \end{array}\)

This is one is obvious . . . with a little Thinking.

How can we have: .a 3-digit number ... subtract a 1-digit number ... and get a 2-digit number?

Suppose the 3-digit number is 200 (or larger?)
Then: .\(\displaystyle 200 - 9 \:=\:191\) . . .a 3-digit number.

Can you see that the 3-digit number must begin with "1" ? .\(\displaystyle A = 1\)

\(\displaystyle \text{We have: } \;\begin {array}{c}1\;B\;B \\ - \quad 1 \\ \hline \;\;D\;D \end{array}\)


\(\displaystyle \text{Suppose }\,B = 2\)
\(\displaystyle \text{Then we have: }\:122 - 1 \:=\:121\) . . . a 3-digit number.

\(\displaystyle \text{The only solution is: }\;\begin{array}{c}1 \:0\:0 \\ - \;\;\; 1 \\ \hline \;\;\;9\:9\end{array}\)

\(\displaystyle 2)\;\;\begin{array}{c}\quad E\;E \\ +\;F\;F \\ \hline E\;E\;K\end{array}\)

We add two 2-digit numbers and get a 3-digit sum.
Can you see that the 3-digit number must begin with "1"?

\(\displaystyle \text{So we have: }\;\begin{array}{c}\quad 1\;1 \\ +\;F\;F \\ \hline 1\;\;1\;\;K\end{array}\)

\(\displaystyle \text{If }F = 8,}\text{ we have: }\:11 + 88 \:=\:99,\,\text{ a 2-digit number.}\)

\(\displaystyle \text{Hence: }\ = 9\)

. . \(\displaystyle \text{and we have: }\;\begin{array}{c}\quad 1\;1 \\ +\;9\;9 \\ \hline \;1\:1\;0\end{array}\)

\(\displaystyle 3)\;\;\begin{array}{c}\;\;\;P\;A \\ +\;A\;P \\ \hline P\;U\;G \end{array} \quad\Leftarrow \quad\text{This problem has }no\text{ solution!}\)

Two 2-digit numbers add up to a 3-digit number.
\(\displaystyle \text{Can you see that the 3-digit number must begin with "1"? }\ = 1\)

\(\displaystyle \text{So we have: }\;\begin{array}{c}\;\;\;1\;A \\ +\;A\;1 \\ \hline 1\;U\;G \end{array}\)

\(\displaystyle \text{Suppose }A = 8\)
. . \(\displaystyle \text{Then we have: }\;\begin{array}{c}\;\;\;1\;8 \\ +\;8\;1 \\ \hline \;\;\;9\;9 \end{array}\begin{array}{c}\\ \\ \leftarrow\text{ a 2-digit sum}\end{array}\)

\(\displaystyle \text{Hence: }\:A = 9\)

. . \(\displaystyle \begin{array}{c}\;\;\;1\;9 \\ +\;9\;1 \\ \hline 1\;1\;0 \end{array}\)

\(\displaystyle \text{But this means that: }\:U = 1\)

\(\displaystyle \text{If each letter represents a }di\!f\!\!f\!erent\text{ digit, this problem has no solution.}\)

 

[wrb]

Hi other of the two guys who does homework

Thank you for your explanation. Are there several other possibilities or can you prove that there is no others in any othe the three problems.